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anonymous

5 years ago

the definition states " two sets S and T are called equinumerous, and we write S ~ T, if there exists a bijective function from S onto T". Bijective definition is "a function is bijective if it is surjective and injective". I'm asked to "Prove that if (S \ T) is equinumerous to (T \ S), then S is equinumerous to T" how do i do that?

More answers

anonymous

5 years ago

set S without T

KingGeorge

5 years ago

So there are the same number of elements in S that aren't in T as there are elements in T that aren't in S. Let \(U=S\cap T\). So \(S\setminus T=S\setminus (S\cap T)\). Then \(S=S\setminus T +U\).
Now let \(f:S\setminus T\to T\setminus S\) such that \(f\) is a bijection. Now define\[g:S\setminus T+U\longrightarrow T\setminus S+U\]by\[
g(s)=\begin{cases}
f(s)\qquad s\in S\setminus T \\
s\qquad \;\;\;\;\,s\in U
\end{cases}\]

KingGeorge

5 years ago

Note that if we restrict \(g\) to only the domain \(U\), we get a bijective function since \(g(g(s))=s\). I.e., \(g\) has an inverse.
Similarly, if we restrict \(g\) to only the domain \(S\setminus T\), then we have a bijective function since \(f\) is bijective. So \(g\) is bijective over its whole domain, and is therefore a bijective function.
Finally, since \(S\setminus T+U=S\) and \(T\setminus S+U=T\), \(S\) and \(T\) are equinumerous.